Question

A wire with a resistance of 5.6 Ω is drawn out through a die so that its new length is 3 times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

Answer 1

Given that density of material and resisivity of wire is unchanged, So Volume of wire will also remain constant.

Volume = Crosse-section Area*Length

V = A*L

Since Volume is constant, So

A1*L1 = A2*L2

A2 = A1*(L1/L2)

Given that new length is three times the original length, So L1/L2 = 1/3

A2 = A1*(1/3) = A1/3

A1/A2 = 3

Now Resistance of a wire is given by:

R = rho*L/A, Since resistivity remains constant, So

R1/R2 = (L1/L2)*(A2/A1)

R2 = Resistance of new wire = R1*(L2/L1)*(A1/A2)

R2 = R1*(3)*(3)

Since R1 = 5.6 ohm, So

R2 = 5.6*3*3

R2 = 50.4 ohm = resistance of longer wire